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3 years ago

INEED THIS GRADUATE PLEASE HELP

Mathematics

1 answer:

Rom4ik [11]3 years ago

3 0

Hi, What grade is this?

I am currently doing flvs too and could help you out.

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Step-by-step explanation:

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2 years ago

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What is the Pythagorean Theorem

Vlad [161]

Answer:

see below

Step-by-step explanation:

The Pythagorean theorem is for right triangles. It uses the two legs and the hypotenuse

a^2 + b^2 = c^2 where a and b are the legs and c is the hypotenuse

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2 years ago

(1/1+sintheta)=sec^2theta-secthetatantheta pls help me verify this

Xelga [282]

Answer:

See Below.

Step-by-step explanation:

We want to verify the equation:

$\displaystyle \frac{1}{1+\sin\theta} = \sec^2\theta - \sec\theta \tan\theta$

To start, we can multiply the fraction by (1 - sin(θ)). This yields:

$\displaystyle \frac{1}{1+\sin\theta}\left(\frac{1-\sin\theta}{1-\sin\theta}\right) = \sec^2\theta - \sec\theta \tan\theta$

Simplify. The denominator uses the difference of two squares pattern:

$\displaystyle \frac{1-\sin\theta}{\underbrace{1-\sin^2\theta}_{(a+b)(a-b)=a^2-b^2}} = \sec^2\theta - \sec\theta \tan\theta$

Recall that sin²(θ) + cos²(θ) = 1. Hence, cos²(θ) = 1 - sin²(θ). Substitute:

$\displaystyle \displaystyle \frac{1-\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta \tan\theta$

Split into two separate fractions:

$\displaystyle \frac{1}{\cos^2\theta} -\frac{\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta\tan\theta$

Rewrite the two fractions:

$\displaystyle \left(\frac{1}{\cos\theta}\right)^2-\frac{\sin\theta}{\cos\theta}\cdot \frac{1}{\cos\theta}=\sec^2\theta - \sec\theta \tan\theta$

By definition, 1 / cos(θ) = sec(θ) and sin(θ)/cos(θ) = tan(θ). Hence:

$\displaystyle \sec^2\theta - \sec\theta\tan\theta \stackrel{\checkmark}{=} \sec^2\theta - \sec\theta\tan\theta$

Hence verified.

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2 years ago

Reflection over the x axis ;vertical shift down 7 units

pochemuha

I need a picture to find the answer

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3 years ago

G is the incenter, or point of concurrency, of the angle bisectors of ΔACE.

vova2212 [387]

Answer:

▢ $\overline{DG}$ ≅ $\overline{FG}$

▢ $\overline{DG}$ ≅ $\overline{BG}$

▢ $\overline{GE}$ bisects $\displaystyle ∠DEF$

▢ $\overline{GA}$ bisects $\displaystyle ∠BAF$

Step-by-step explanation:

As you can see, angles $\displaystyle BAF$ and $\displaystyle DEF$ form two half-kites, which you should know have two pairs of congruent sides, therefore segments $\displaystyle GA$ and $\displaystyle GE$ are what you call angle bisectours, which is why segments $\displaystyle FG,$ $\displaystyle DG,$ and $\displaystyle BG$ are congruent.

I am joyous to assist you at any time.

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2 years ago